Troupe 2020 divisor sums representable as sum two

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theorem_1_2: Proves that the number of n <= x for which the sum of proper divisors s(n) is a sum of two squares is bounded above and below by absolute constant multiples of x/(log x)^{1/2}.

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Troupe, Lee, Divisor sums representable as the sum of two squares. Proc. Amer. Math. Soc. 148 (2020), no. 10, 4189--4202, DOI 10.1090/proc/15104. The copy read for this card is arXiv:1902.11171v1 (28 February 2019, 14 pages), whose labels the card uses. The arXiv record names arXiv's non-exclusive distribution license (arXiv:1902.11171), every other right reserved.

Theorem 1.2 (p. 1) proves that B_s(x), the number of n <= x for which s(n) (the sum of the proper divisors of n) is a sum of two squares, satisfies B_s(x) ≍ x/(log x)^{1/2} with absolute implied constants, the same order of magnitude as Landau's count B(x) ~ C x/(log x)^{1/2} of integers up to x that are themselves sums of two squares. This confirms, in a strong quantitative form, the special case A = {sums of two squares} of the Erdos-Granville-Pomerance-Spiro conjecture (Conjecture 1.1) that s^{-1}(A) has density zero whenever A does; earlier special cases were known for A the primes, for palindromes, and for sets with counting function << x^{1/2 + eps(x)} for a fixed eps(x) tending to 0. The proof restricts to n outside E(x) = {n <= x : P(n) <= x^{1/log log x} or P(n)^2 | n}, P(n) the largest prime factor of n, shown by Lemma 2.1 to have size << x/(log x)^2 for sufficiently large x using de Bruijn's smooth-number bound (Proposition 2.2), and then exploits the representation n = mP with s(n) = P s(m) + sigma(m). Section 3 (Proposition 3.1) shows that all but o(x/(log x)^{1/2}) of the counted n satisfy three conditions on m; Brun's sieve and a Mertens theorem in arithmetic progressions modulo 4 (Theorem 2.3) give the upper bound in Section 4, and a theorem of Friedlander and Iwaniec (Theorem 5.1, their Opera de Cribro Theorem 14.8) gives the lower bound in Section 5. The upper bound alone gives the case A = {sums of two squares} of Erdos problem 955.

Read status: claims checked for Theorem 1.2 and the Section 2 statements listed below against the print (pp. 1--5); the proofs of Sections 3--5 (pp. 5--13) were read for structure only. Result page: theorem_1_2.

Source: https://arxiv.org/abs/1902.11171.

Bears on. #955: Theorem 1.2 gives B_s(x) = O(x/(log x)^{1/2}) = o(x), so the preimage under s of the sums of two squares, a density-zero set by Landau's theorem, has density zero; this is one case of the problem's assertion, and the paper treats no other target set.

Results to transcribe.

• Theorem 1.2 (p. 1): B_s(x), the count of n <= x with s(n) a sum of two squares, satisfies B_s(x) ≍ x/(log x)^{1/2} with absolute implied constants.
• Conjecture 1.1 (p. 1): Erdős-Granville-Pomerance-Spiro: if A has asymptotic density zero then so does s^{-1}(A); Theorem 1.2 confirms the sum-of-two-squares case quantitatively.
• Lemma 2.1 (p. 2): For sufficiently large x, the exceptional set E(x) = {n <= x : P(n) <= x^{1/log log x} or P(n)^2 | n} has size << x/(log x)^2.
• Proposition 2.2 (p. 2): De Bruijn's smooth-number bound: if x >= y >= 2 and (log x)^2 <= y <= x, then Psi(x, y) <= x / u^{u + o(u)} as u = log x / log y tends to infinity; used to bound the smooth part of E(x).
• Theorem 2.3 (p. 2): Mertens's theorem generalized to arithmetic progressions: for fixed integers a, b, the sum of 1/p over primes p <= x with p = a (mod b) is (1/phi(b)) log log x + c_{a,b} + O_b(1/log x); applied with modulus 4 and residues 1 and 3.

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